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Suppose $0 < a < b$, and let GM and AM be respectively the geometric and arithmetic means of $a$ and $b$. Does the mapping $(a,b) \mapsto (\mathrm{GM}, \mathrm{AM})$ have a well-behaved compositional $n$th root? In what senses might such an $n$th root be unique?

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Although it's not clear to me how to raise this operation to fractional powers, it's easy to raise it to negative powers (as long as we insist that the first component in the ordered pair be the smaller one---so I suppose we should probably just work with _un_ordered pairs). $$ a_{-1} = b_0 - \sqrt{b_0^2 - a_0^2},\qquad b_{-1} = b_0 + \sqrt{b_0^2 - a_0^2} $$ – Michael Hardy Aug 2 2010 at 2:07
Here's what made me think of this: An exercise in a book challenges the reader to show that if $c, d$ are respectively the geometric and arithmetic means of $a, b$, then $$ \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + c^2)(x^2 + d^2)}} = \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2 + b^2)}}. $$ Writing the first integral as $$ \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + c^2)(x^2 + d^2)}} $$ and then $$ u = \frac{x - \frac{1b}{x^2}}{2} $$ does it. That shows this works if one does the operation once. Induction shows it works if done $n$ for integer $n$. So what about non-integers? – Michael Hardy Aug 2 2010 at 23:56
I meant: Writing the first integral as $$ \int_{-\infty}^\infty \frac{du}{\sqrt{(u^2 + c^2)(u^2 + d^2)}} $$ and $$ u = \frac{x - \frac{ab}{x^2}}{2}. $$ – Michael Hardy Aug 2 2010 at 23:58
Haste makes waste. One more try: $$ u = \frac{x - \frac{ab}{x}}{2} $$ – Michael Hardy Aug 2 2010 at 23:59
I see that great prominence is given to this integral in the first chapter of Borwein & Borwein's Pi and the AGM. Of course it follows from what I said above that $$ \int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2 + b^2)}} = \frac{\pi}{\text{AGM}(a,b)}. $$ – Michael Hardy Aug 4 2010 at 15:57

3 Answers

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Here is a different (partial) answer, which does not assume any regularity of the square root $G$ of $F$ along the diagonal $a=b$. Let $M(a,b)$ denote the arithmetico-geometric mean. We have $F^{(n)}(a,b)\rightarrow(M(a,b),M(a,b))$ as $n\rightarrow+\infty$, and $M=M\circ F$. The sector $0< a< b$ is foliated by the level curves $\gamma_t=M^{-1}(t)$. By homogeneity, $\gamma_t=t\gamma_1$. Each of these curves is invariant under $F$. It is therefore natural to look for a square root $G$ that preserves every $\gamma_t$. It is enough to construct $G$ over $\gamma_1$, and then to extend it to $\gamma_t$ by homogenity: $G(ta,tb)=tG(a,b)$.

The curve $\gamma_1$ is transversal to the rays, and therefore can be parametrized by the angle $\theta\in(0,\pi/4)$. Its end point at $\pi/4$ is $(1,1)$. The restriction of $F$ over $\gamma_1$ is thus conjugated to a map $f:(0,\pi/4)\rightarrow(0,\pi/4)$. I is not hard to see that $f(\theta)> \theta$, because $$\frac{a+b}{2\sqrt{ab}}< \frac{b}{a}.$$ Likewise, $f'> 0$.

There remains to find a $g:(0,\pi/4)\rightarrow(0,\pi/4)$, such that $g\circ g=f$.

A construction of $g$ could be made as follow. First find a vector field $X$ over $(0,\pi/4)$, whose flow at time $1$ is $f$. This is the hard part, for which I shall ask MO. Then set $g$ the flow of $X$ at $t=1/2$. One obtains also the $n$-th roots by taking the flow of $X$ at time $1/n$.

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The flow, you are searching for, does exist. In order to show this, take a point $x_0$ \in $(0,\pi/4)$. Let $x_n$ denote $f^n(x_0)$, where power is taken with respect to composition. Define some monotone smooth map $\phi\colon [x_0, x_1]\to[0,1]$, such that $\phi(x_0)=0$, $\phi(x_1)=1$ and $\phi^{(n)}(x_1)=(\partial/\partial x)|_{x=x_0}(\phi(f(x))$ for $n\geq1$. Then let $\phi(x)=\phi(f^n(x))-n$ where $n\in\mathbb{N}$ is such that $f^n(x)\in [x_0,x_1]$. The vector field you are interested in is just a pullback of $\partial/\partial x$ with respect to $\phi$. – Fiktor Sep 24 2010 at 22:19
I'm sorry, one should take $n\in \mathbb{Z}$, not necessarily in $\mathbb{N}$. – Fiktor Sep 24 2010 at 22:21
The hard part mentionned in my answer is OK. See Pietro Majer's answer to my MO question mathoverflow.net/questions/39907/… . Therefore the square root, and also the $n$-roots of $F$ do exist. – Denis Serre Sep 25 2010 at 6:51
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Your mapping fixes the diagonal: $F(a,a)\equiv(a,a)$, so I presume that you are interested in a root $G$ that also fixes the diagonal. Under this natural restriction, there does not exist a twice differentiable square root $G$.

Proof. Let $m$ be a fixed point of $F$, hence of $G$. Then $DF(m)=(DG(m))^2$ is the matrix $A$ equal to $$\begin{pmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{pmatrix}.$$ Its only square roots are $\pm A$. But since $G$ fixes the diagonal, $DG(m)$ has an eigenvalue $1$, and therefore $DG(m)=A$.

Now, expanding at second order the identity $F(m+h)=G(G(m+h))$, and using the previous result, one obtains $AD^2G(m)=D^2F(m)$ at fixed points. But since $A$ is rank one, this implies that $h\mapsto D^2F(m)h\otimes h$ is not onto. Specifically, since $(1,-1)A=0$, one should have $D^2f(m)h\otimes h=0$ for every fixed point $m=(a,a)$ and increment $h$, where $f:=GM-AM$. But this is false, instead we have $$D^2f(m)h\otimes h=-\frac{1}{8a}(h_2-h_1)^2.$$ QED

The proof applies to $n$th roots instead of square roots. If $G$ is an $n$th root, one still has $DG(m)=A$ at fixed point. From this, it follows that $$G^{(n)}(m+h)=m+Ah+\frac{1}{2}AD^2G(m)h\otimes h,$$ independently of $n$. For this calculation, it is important to notice that $m+Ah$ is a fixed point too, and that $A^2=A$. Hence the same conclusion: there does not exist a smooth $n$th root of $F$ fixing the diagonal points.

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@Michael. If you think that this answer solves the problem, please feel free to accept it. – Denis Serre Sep 22 2010 at 15:18
The lack of answer from you suggests to me that my answer is not acceptable. Perhaps because your domain is $0< a< b$ and therefore you don't care about the behaviour of the square root at the boundary $a=b$. My answer uses regularity there, and you might expect a square root that is not regular, at least not twice differentiable at the boundary. – Denis Serre Sep 24 2010 at 20:40
Hello. I haven't yet looked this over closely; I've been busy with a few things. I'll get to it. – Michael Hardy Sep 25 2010 at 17:42
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The following is more a sidenote than an answer to the problem of the fractional iteration of this map, however it introduces a connection to the half-iterate of the exponential/logarithm-map, symmetrizes the iteration and maybe interesting for that.

Define the two functions for the half-exponential and half-logarithm
$ h(h(x)) = \exp(x) $
$ g(g(x)) = \log(x) $

It des not matter which base for the exponentiation we use, say we use base $ b=\sqrt(2) $ which allows a real-valued solution for the half-iterate of $ b^x $ and $ log_b(x) $ (I've checked the process using that base and applying "regular fractional iteration")

Then in the original iterated map

$ a_{k+1} = \frac {a_k+b_k} 2 $
$ b_{k+1} = \sqrt {a_k * b_k} $

substitute the initial $a_0 $ by $ A_0 = h(a_0) $ and $b_0$ by $ B_0 = h(b_0) $
Then define the iteration

$ A_{k+1} = g(\frac {h (A_k) + h (B_k)} {2}) $

$ B_{k+1} = h(\frac {g (A_k)+g (B_k)} {2}) $

Then $ g(A_{oo}) $ and $ g(B_{oo}) $ give the AGM(a,b).

I do not see at the moment how this could be improved to allow a fractional iteration, but perhaps the idea suggests a viable direction.

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